Integrate. $ \int -10\sin(x)\,dx $ $=$ $+ C$
Answer: We need a function whose derivative is $-10\sin(x)$. We know that the derivative of $\cos(x)$ is $-\sin(x)$, so let's start there: $\dfrac{d}{dx} \cos(x) = -\sin(x)$ Now let's multiply by $10$ : $\dfrac{d}{dx}\left[ 10\cos(x) \right]= -10\sin(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int -10\sin(x)\,dx =10 \cos(x)\, + C$ The answer: $10 \cos(x)\, + C$